原始 URL.createObjectURL 正在产生 UUID. 您可以利用此。

function uuid() {
  const url = URL.createObjectURL(new Blob())
  const [id] = url.toString().split('/').reverse()
  URL.revokeObjectURL(url)
  return id
}

这可能对某人有用......

var d = new Date().valueOf();
var n = d.toString();
var result = '';
var length = 32;
var p = 0;
var chars = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';

for (var i = length; i > 0; --i){
    result += ((i & 1) && n.charAt(p) ? '<b>' + n.charAt(p) + '</b>' : chars[Math.floor(Math.random() * chars.length)]);
    if(i & 1) p++;
};

HTTPS://jsfiddle.net/j0evrdf1/1/

这里有很多正确的答案,但不幸的是,包含的代码样本是相当密码和难以理解的。

请注意,随后的代码使用二进制字母以提高可读性,因此需要ECMAScript 6。

Node.js 版本

function uuid4() {
  let array = new Uint8Array(16)
  crypto.randomFillSync(array)

  // Manipulate the 9th byte
  array[8] &= 0b00111111 // Clear the first two bits
  array[8] |= 0b10000000 // Set the first two bits to 10

  // Manipulate the 7th byte
  array[6] &= 0b00001111 // Clear the first four bits
  array[6] |= 0b01000000 // Set the first four bits to 0100

  const pattern = "XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX"
  let idx = 0

  return pattern.replace(
    /XX/g,
    () => array[idx++].toString(16).padStart(2, "0"), // padStart ensures a leading zero, if needed
  )
}

浏览器版本

只有第二条线是不同的。

function uuid4() {
  let array = new Uint8Array(16)
  crypto.getRandomValues(array)

  // Manipulate the 9th byte
  array[8] &= 0b00111111 // Clear the first two bits
  array[8] |= 0b10000000 // Set the first two bits to 10

  // Manipulate the 7th byte
  array[6] &= 0b00001111 // Clear the first four bits
  array[6] |= 0b01000000 // Set the first four bits to 0100

  const pattern = "XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX"
  let idx = 0

  return pattern.replace(
    /XX/g,
    () => array[idx++].toString(16).padStart(2, "0"), // padStart ensures a leading zero, if needed
  )
}

测试

最后,相应的测试(Jasmine)。

describe(".uuid4()", function() {
  it("returns a UUIDv4 string", function() {
    const uuidPattern = "XXXXXXXX-XXXX-4XXX-YXXX-XXXXXXXXXXXX"
    const uuidPatternRx = new RegExp(uuidPattern.
      replaceAll("X", "[0-9a-f]").
      replaceAll("Y", "[89ab]"))

    for (let attempt = 0; attempt < 1000; attempt++) {
      let retval = uuid4()
      expect(retval.length).toEqual(36)
      expect(retval).toMatch(uuidPatternRx)
    }
  })
})

UUID v4 解释

非常好的解释 UUID 版本 4 在这里: 创建一个符合 RFC 4122 的 UUID。

最后笔记

此外,有很多第三方包. 但是,只要你只是基本的需求,我不推荐它们. 事实上,没有很多赢得和几乎失去。 作者可以追求最薄的比特的性能,“修复”的事情,不应该固定,当涉及到安全,这是一个危险的想法. 同样,他们可能会引入其他错误或不一致性。

下面是一個功能,從一條線或一條偶然的 UUID 生成一個静態 UUID 如果沒有連線提供:

函数 stringToUUID(str) {如果(str === 未定义 <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk>

吉普赛

GitHub上的JavaScript项目 - https://github.com/LiosK/UUID.js

参见 RFC 4122 http://www.ietf.org/rfc/rfc4122.txt. 功能 产生 RFC 4122 符合 UUIDs. 版本 4 UUIDs (UUIDs 从随机数字) 和版本 1 UUIDs (基于时间的 UUIDs) 可用. UUID 对象允许各种访问 UUID 包括访问 UUID 字段。

我想了解布罗法的答案,所以我扩展了它并添加了评论:

var uuid = function () {
    return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(
        /[xy]/g,
        function (match) {
            /*
            * Create a random nibble. The two clever bits of this code:
            *
            * - Bitwise operations will truncate floating point numbers
            * - For a bitwise OR of any x, x | 0 = x
            *
            * So:
            *
            * Math.random * 16
            *
            * creates a random floating point number
            * between 0 (inclusive) and 16 (exclusive) and
            *
            * | 0
            *
            * truncates the floating point number into an integer.
            */
            var randomNibble = Math.random() * 16 | 0;

            /*
            * Resolves the variant field. If the variant field (delineated
            * as y in the initial string) is matched, the nibble must
            * match the mask (where x is a do-not-care bit):
            *
            * 10xx
            *
            * This is achieved by performing the following operations in
            * sequence (where x is an intermediate result):
            *
            * - x & 0x3, which is equivalent to x % 3
            * - x | 0x8, which is equivalent to x + 8
            *
            * This results in a nibble between 8 inclusive and 11 exclusive,
            * (or 1000 and 1011 in binary), all of which satisfy the variant
            * field mask above.
            */
            var nibble = (match == 'y') ?
                (randomNibble & 0x3 | 0x8) :
                randomNibble;

            /*
            * Ensure the nibble integer is encoded as base 16 (hexadecimal).
            */
            return nibble.toString(16);
        }
    );
};